Simplify; express your answer in exponential form. Assume $q\neq 0, y\neq 0$. $\dfrac{{(q^{3})^{5}}}{{(q^{3}y^{5})^{-1}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(q^{3})^{5} = q^{15}}$ In the denominator, we can use the distributive property of exponents. ${(q^{3}y^{5})^{-1} = (q^{3})^{-1}(y^{5})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{3})^{5}}}{{(q^{3}y^{5})^{-1}}} = \dfrac{{q^{15}}}{{q^{-3}y^{-5}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{15}}}{{q^{-3}y^{-5}}} = \dfrac{{q^{15}}}{{q^{-3}}} \cdot \dfrac{{1}}{{y^{-5}}} = q^{{15} - {(-3)}} \cdot y^{- {(-5)}} = q^{18}y^{5}$.